= Solution
The highest-field plateau is labelled $R_H/(h/e^2)=1$, so it is the \b[$\nu=1$ plateau] in the <filling factor>, with the longitudinal resistivity returning to approximately zero after the transition near \b[10 T]. The odd integer indicates a resolved spin branch at these fields; assuming a permanent twofold <spin degeneracy> for every orbital <Landau level> would misidentify the plotted sequence.
For a nominal exactly filled $\nu=1$ state at a chosen field $B_*$, the density target is $n_*=eB_*/h$. Combining this with the same gate capacitor model gives the required gate-voltage change relative to the original ungated-density reference:
$$
\boxed{\Delta V_g(B_*)=\frac{ed}{\epsilon_0\epsilon_r}
\left(\frac{eB_*}{h}-n_s\right).}
$$
Taking $B_*\simeq10\ \mathrm T$ as the graph's onset-scale estimate gives $n_*=2.418\times10^{15}\ \mathrm{m^{-2}}$ and
$$
\boxed{\Delta V_g\simeq-0.0806\ \mathrm V.}
$$
Relative to the positive 0.2 V setting in part (v), this is an additional change of approximately \b[$-0.281\ \mathrm V$]. An absolute laboratory gate voltage cannot be obtained without its initial offset or threshold; the calculated quantities are voltage changes.
There is a graphical/model qualification: a <quantum Hall plateau> occupies a finite field interval, and its onset is not an exact statement that $n_s=eB/h$ at that edge. The plot identifies $\nu=1$ from about 10 T onward but does not specify a unique plateau-centre field or the gate setting at which the trace was recorded. The boxed formula gives the result for any selected target field. For comparison, the nominal exact-filling fields are 12.39 T at the original density and 18.34 T at the unchanged positive 0.2 V density. Thus the graph must not be combined silently with the latter density as if its $\nu=1$ onset uniquely fixed the same gate bias.
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